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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Chain of floors
Assassino9931   1
N 13 minutes ago by pi_quadrat_sechstel
Source: Vojtech Jarnik IMC 2025, Category I, P2
Determine all real numbers $x>1$ such that
\[ \left\lfloor\frac{n+1}{x}\right\rfloor = n - \left\lfloor \frac{n}{x} \right\rfloor + \left \lfloor \frac{\left \lfloor \frac{n}{x} \right\rfloor}{x}\right \rfloor - \left \lfloor \frac{\left \lfloor \frac{\left\lfloor \frac{n}{x} \right\rfloor}{x} \right\rfloor}{x}\right \rfloor + \cdots \]for any positive integer $n$.
1 reply
Assassino9931
May 2, 2025
pi_quadrat_sechstel
13 minutes ago
2015 Taiwan TST Round 2 Quiz 1 Problem 2
wanwan4343   8
N 19 minutes ago by Want-to-study-in-NTU-MATH
Source: 2015 Taiwan TST Round 2 Quiz 1 Problem 2
Let $\omega$ be the incircle of triangle $ABC$ and $\omega$ touches $BC$ at $D$. $AD$ meets $\omega$ again at $L$. Let $K$ be $A$-excenter, and $M,N$ be the midpoint of $BC,KM$, respectively. Prove that $B,C,N,L$ are concyclic.
8 replies
wanwan4343
Jul 12, 2015
Want-to-study-in-NTU-MATH
19 minutes ago
Inspired by Butterfly
sqing   2
N 20 minutes ago by sqing
Source: Own
Let $ a,b,c>0. $ Prove that
$$a^2+b^2+c^2+ab+bc+ca+abc-3(a+b+c) \geq 34-14\sqrt 7$$$$a^2+b^2+c^2+ab+bc+ca+abc-\frac{433}{125}(a+b+c) \geq \frac{2(57475-933\sqrt{4665})}{3125} $$
2 replies
sqing
Today at 8:52 AM
sqing
20 minutes ago
Triangle form by perpendicular bisector
psi241   52
N 23 minutes ago by ihatemath123
Source: IMO Shortlist 2018 G5
Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. A line $\ell$ intersects the lines $AI$, $BI$, and $CI$ at points $D$, $E$, and $F$, respectively, distinct from the points $A$, $B$, $C$, and $I$. The perpendicular bisectors $x$, $y$, and $z$ of the segments $AD$, $BE$, and $CF$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$.
52 replies
psi241
Jul 17, 2019
ihatemath123
23 minutes ago
Concentric Circles
MithsApprentice   63
N an hour ago by QueenArwen
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
63 replies
MithsApprentice
Oct 9, 2005
QueenArwen
an hour ago
D is incenter
Layaliya   7
N an hour ago by rong2020
Source: From my friend in Indonesia
Given an acute triangle \( ABC \) where \( AB > AC \). Point \( O \) is the circumcenter of triangle \( ABC \), and \( P \) is the projection of point \( A \) onto line \( BC \). The midpoints of \( BC \), \( CA \), and \( AB \) are \( D \), \( E \), and \( F \), respectively. The line \( AO \) intersects \( DE \) and \( DF \) at points \( Q \) and \( R \), respectively. Prove that \( D \) is the incenter of triangle \( PQR \).
7 replies
Layaliya
Apr 3, 2025
rong2020
an hour ago
collinear points in the regular 7-gon
parmenides51   2
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1995 OMM P3
$A, B, C, D$ are consecutive vertices of a regular $7$-gon. $AL$ and $AM$ are tangents to the circle center $C$ radius $CB$. $N$ is the intersection point of $AC$ and $BD$. Show that $L, M, N$ are collinear.
2 replies
parmenides51
Jul 28, 2018
FrancoGiosefAG
an hour ago
geometry
Wendyyy_23   5
N an hour ago by rong2020

The inscribed circle of triangle ABC is tangent to the sides BC, AC, AB respectively at P, K, N. The line BK intersects the circle with the inscribed triangle ABC at L. T is the intersection of AL and KN, Q is the intersection of CL and KP. Prove that the lines BK, NQ, PT are concurrent
5 replies
Wendyyy_23
Jun 10, 2020
rong2020
an hour ago
Show that TA=TM
goldeneagle   59
N 3 hours ago by Siddharthmaybe
Source: Iran TST 2011 - Day 1 - Problem 1
In acute triangle $ABC$ angle $B$ is greater than$C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$ and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$.
59 replies
goldeneagle
May 10, 2011
Siddharthmaybe
3 hours ago
Who loves config geo with orthocenter?
Assassino9931   11
N 4 hours ago by ravengsd
Source: RMM 2024 Shortlist G2
Let $ABC$ be an acute triangle with orthocentre $H$ and circumcircle $\Gamma$. Let $D$ be the point
diametrically opposite $A$ on $\Gamma$. The line through $H$, parallel to $BC$, intersects $AB$ and $AC$ at $X$
and $Y$, respectively. Let $AD$ intersect the circumcircle of triangle $DXY$ again at $S$. Let the tangent to $\Gamma$ at $A$ intersect $XY$ at $T$. Prove that lines $DT$ and $HS$ intersect on $\Gamma$.
11 replies
Assassino9931
Feb 17, 2025
ravengsd
4 hours ago
Geometry
MathsII-enjoy   4
N Today at 7:51 AM by whwlqkd
Given triangle $ABC$ inscribed in $(O)$ with $M$ being the midpoint of $BC$. The tangents at $B, C$ of $(O)$ intersect at $D$. Let $N$ be the projection of $O$ onto $AD$. On the perpendicular bisector of $BC$, take a point $K$ that is not on $(O)$ and different from M. Circle $(KBC)$ intersects $AK$ at $F$. Lines $NF$ and $AM$ intersect at $E$. Prove that $AEF$ is an isosceles triangle.
4 replies
MathsII-enjoy
May 15, 2025
whwlqkd
Today at 7:51 AM
Tangencies with cyclic quadrilateral
tapir1729   21
N Today at 4:31 AM by Mathandski
Source: TSTST 2024, problem 4
Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to $\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$. Show that $OP=OQ$.

Merlijn Staps
21 replies
tapir1729
Jun 24, 2024
Mathandski
Today at 4:31 AM
Angles in a triangle with integer cotangents
Stear14   0
Today at 3:16 AM
In a triangle $ABC$, the point $M$ is the midpoint of $BC$ and $N$ is a point on the side $BC$ such that $BN:NC=2:1$. The cotangents of the angles $\angle BAM$, $\angle MAN$, and $\angle NAC$ are positive integers $k,m,n$.
(a) Show that the cotangent of the angle $\angle BAC$ is also an integer and equals $m-k-n$.
(b) Show that there are infinitely many possible triples $(k,m,n)$, some of which consisting of Fibonacci numbers.
0 replies
Stear14
Today at 3:16 AM
0 replies
An important lemma of isogonal conjugate points
buratinogigle   6
N Today at 1:26 AM by buratinogigle
Source: Own
Let $P$ and $Q$ be two isogonal conjugate with respect to triangle $ABC$. Let $S$ and $T$ be two points lying on the circle $(PBC)$ such that $PS$ and $PT$ are perpendicular and parallel to bisector of $\angle BAC$, respectively. Prove that $QS$ and $QT$ bisect two arcs $BC$ containing $A$ and not containing $A$, respectively, of $(ABC)$.
6 replies
buratinogigle
Mar 23, 2025
buratinogigle
Today at 1:26 AM
Weighted graph problem
egxa   1
N Apr 21, 2025 by internationalnick123456
Source: All Russian 2025 10.4
In the plane, $10^6$ points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than $1$. For every group of $6$ marked points, he calculated the sum of the numbers on all $15$ connecting segments. It turned out that the absolute value of each such sum is at least \(C\), and there are both positive and negative such sums. What is the maximum possible value of \(C\)?
1 reply
egxa
Apr 18, 2025
internationalnick123456
Apr 21, 2025
Weighted graph problem
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 10.4
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egxa
211 posts
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In the plane, $10^6$ points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than $1$. For every group of $6$ marked points, he calculated the sum of the numbers on all $15$ connecting segments. It turned out that the absolute value of each such sum is at least \(C\), and there are both positive and negative such sums. What is the maximum possible value of \(C\)?
This post has been edited 2 times. Last edited by egxa, Apr 21, 2025, 7:22 PM
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internationalnick123456
135 posts
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Claim. There exists $7$ points such that the sums corr. to $6$ among these points attain both positive and negative values.
Proof. Notice that from any group of $6$ points, we can replace one point at a time with another from the remaining point, and after finitely many steps, we can obtain any group of $6$ points. By continuity, we conclude our claim.
Now, denote the $7$ points as $A_1, A_2, \ldots, A_7$, and let $S_i$ be the sum corresponding to the subset of $6$ points excluding $A_i$.
Wlog, assume $S_1, \ldots, S_k > 0$ and $S_{k+1}, \ldots, S_7 < 0$.
Let $X,Y,Z$ be the sum over all segments $A_iA_j$ with: $1\leq i<j\leq k, 1\leq i\leq k<j\leq 7, k<i<j\leq 7$, respectively.
Then we have $$kC\leq S(A_1)+\cdots +S(A_k)= (k-2)X + (k-1)Y + kZ\quad(1)$$$$-(7-k)C\geq S(A_{k+1})+\cdots + S(A_7)=(7-k)X + (6-k)Y + (5-k)Z\quad(2)$$
Considering $(7-k)\times (1) - (k-2)\times (2)$, we get $$(7-k)(2k-2)C\leq 5Y + 10Z\leq 5k(7-k)+5(7-k)(6-k)=30(7-k)\Rightarrow C\leq \dfrac{15}{k-1}$$By switching the roles of positive and negative, we also get $C\leq \dfrac{15}{6-k}$
Moreover, considering $(6-k)\times (1) - (k-1)\times (2)$ we have $C(14k-2k^2-7)\leq 5(Z-X)\leq 5(21-7k + k^2)$.
If $k\leq 2$ or $k\geq 5$ then $C\leq \min\{\dfrac{15}{k-1},\dfrac{15}{6-k}\}\leq \dfrac{15}{4}$.
If $k\in\{3,4\}$ then $C\leq \dfrac{45}{17}< \dfrac{15}{4}$.
Hence, in all cases, we conclude $C\leq \dfrac{15}{4}$.
To construct such a configuration that attains the maximum, consider two points $A,B$, and assign $1$ to all segments involving either $A$ or $B$, and $-7/8$ to all remaining segments. In this construction, the sum over any $6$-point subset is exactly $15/4$.
Thus, $C_{\max}=\dfrac{15}{4}$.
This post has been edited 1 time. Last edited by internationalnick123456, Apr 21, 2025, 12:43 PM
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